$f(x, y) = \sin(x) + \cos(y^2)$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2y\sin(y^2)$ (Choice B) B $\cos(x) - 2y\sin(y^2)$ (Choice C) C $\cos(x)$ (Choice D) D $0$
Taking a partial derivative with respect to $y$ means treating $x$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial y} &= \dfrac{\partial}{\partial y} \left[ \sin(x) + \cos({y^2}) \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ \sin(x) \right] + \dfrac{\partial}{\partial y} \left[ \cos({y^2}) \right] \\ \\ &= 0 + {2y} (-\sin({y^2})) \\ \\ &= -{2y}\sin({y^2}) \end{aligned}$ Note that $\dfrac{\partial}{\partial y} \left[ \sin(x) \right] = 0$ since we're treating $x$ as a constant. In conclusion, $\dfrac{\partial f}{\partial y} = -2y\sin(y^2)$